Question

The natural frequency of the system shown in figure is given by $\frac{N}{\pi }\text{Hz}$. Then, the value of $N$ is [Assume pulleys are smooth and massless]

Answer 1

If the mass $1\text{kg}$ is displaced down by $x$ from its mean position, then each pulley will shift down by $x$. Therefore, each spring will extend by $2x$. Hence, tension in the string will be $2kx$.
From figure, net restoring force on mass,
$F=-8kx$
$\Rightarrow ma=-8kx$
$\Rightarrow 1\times a=-8\times 200\times x$
$\Rightarrow \frac{a}{x}=-1600----\left(1\right)$

Now, natural frequency.
$f=\frac{1}{2\pi }\sqrt{\left|\frac{a}{x}\right|}$

$\Rightarrow f=\frac{1}{2\pi }\sqrt{1600}$

$\Rightarrow f=\frac{20}{\pi }=\frac{N}{\pi }$ (given)

$\Rightarrow N=20$